- #1

NZer

- 3

- 0

Im currently revising for my exams and I encountered a problem I hope someone will be able to help me with.

## Homework Statement

Find the volume of the region of space bounded by:

The planes x=0, y=0, z=0, z=3-2x+y and the surface y=1-x^2

## Homework Equations

[tex]\int \int \int _R 1\, dV[/tex]

## The Attempt at a Solution

First I decided to integrate with respect to the z direction as I wouldn't have to worry about splitting up the region yet.

[tex]\int \int \int _0 ^{3-2x+y} 1\, dz\, dy\, dx[/tex]

[tex]= \int \int 3-2x+y\, dy \, dx[/tex]

ok. But now I have a problem due to the surface y=1-x^2 cutting our region defined by the 4 planes. Can we split the region and choose our bounds like below?

[tex]= \int _1 ^{3/2} \int _{1-x^2} ^{2x-3} 3-2x+y\, dy\, dx \; + \int _0 ^1 \int _0 ^{2x-3} 3-2x+y\, dy \, dx[/tex]

Thanks.